Nonlinear Control Lecture # 31 Nonlinear Observers Nonlinear - PowerPoint PPT Presentation

Nonlinear Control Lecture # 31 Nonlinear Observers Nonlinear Control Lecture # 31 Nonlinear Observers

Local Observers x = f ( x, u ) , ˙ y = h ( x ) ˙ x = f (ˆ ˆ x, u ) + H [ y − h (ˆ x )] x = x − ˆ ˜ x ˙ x = f ( x, u ) − f (ˆ ˜ x, u ) − H [ h ( x ) − h (ˆ x )] We seek a local solution for sufficiently small � ˜ x (0) � Linearization at ˜ x = 0 : � ∂f � ∂x ( x ( t ) , u ( t )) − H ∂h ˙ x = ˜ ∂x ( x ( t )) x ˜ Nonlinear Control Lecture # 31 Nonlinear Observers

Steady-state solution: 0 = f ( x ss , u ss ) , 0 = h ( x ss ) Assumption: given ε > 0 , there exist δ 1 > 0 and δ 2 > 0 such that � x (0) − x ss � ≤ δ 1 and � u ( t ) − u ss � ≤ δ 2 ∀ t ≥ 0 ⇒ � x ( t ) − x ss � ≤ ε ∀ t ≥ 0 A = ∂f C = ∂h ∂x ( x ss , u ss ) , ∂x ( x ss ) Assume that ( A, C ) is detectable. Design H such that A − HC is Hurwitz Nonlinear Control Lecture # 31 Nonlinear Observers

Lemma 11.1 For sufficiently small � ˜ x (0) � , � x (0) − x ss � , and sup t ≥ 0 � u ( t ) − u ss � , t →∞ ˜ lim x ( t ) = 0 Proof � 1 ∂f f ( x, u ) − f (ˆ x, u ) = ∂x ( x − σ ˜ x, u ) dσ ˜ x 0 � f ( x, u ) − f (ˆ x, u ) − A ˜ x � = � 1 � � � ∂f x, u ) − ∂f ∂x ( x, u ) + ∂f ∂x ( x, u ) − ∂f � � � ∂x ( x − σ ˜ ∂x ( x ss , u ss ) dσ ˜ x � � � � 0 ≤ L 1 ( 1 2 � ˜ x � + � x − x ss � + � u − u ss � ) � ˜ x � Nonlinear Control Lecture # 31 Nonlinear Observers

x � ≤ L 2 ( 1 � h ( x ) − h (ˆ x ) − C ˜ 2 � ˜ x � + � x − x ss � ) � ˜ x � ˙ x = ( A − HC )˜ ˜ x + ∆( x, u, ˜ x ) x � 2 + k 2 ( ε + δ 2 ) � ˜ � ∆( x, u, ˜ x ) � ≤ k 1 � ˜ x � P ( A − HC ) + ( A − HC ) T P = − I x T P ˜ V = ˜ x x � 2 + c 4 k 1 � ˜ x � 3 + c 4 k 2 ( ε + δ 2 ) � ˜ ˙ x � 2 V ≤ −� ˜ ˙ V ≤ − 1 x � 2 , x � ≤ 1 c 4 k 2 ( ε + δ 2 ) ≤ 1 3 � ˜ for c 4 k 1 � ˜ and 3 3 For sufficiently small � ˜ x (0) � , ε , and δ 2 , the estimation error converges to zero as t tends to infinity Nonlinear Control Lecture # 31 Nonlinear Observers

The Extended Kalman Filter x = f ( x, u ) , ˙ y = h ( x ) ˙ x = f (ˆ ˆ x, u ) + H ( t )[ y − h (ˆ x )] x = x − ˆ ˜ x ˙ ˜ x = f ( x, u ) − f (ˆ x, u ) − H ( t )[ h ( x ) − h (ˆ x )] Expand the right-hand side in a Taylor series about ˜ x = 0 and evaluate the Jacobian matrices along ˆ x ˙ x = [ A ( t ) − H ( t ) C ( t )]˜ ˜ x + ∆(˜ x, x, u ) A ( t ) = ∂f C ( t ) = ∂h ∂x (ˆ x ( t ) , u ( t )) , ∂x (ˆ x ( t )) Nonlinear Control Lecture # 31 Nonlinear Observers

Kalman Filter Gain: H ( t ) = P ( t ) C T ( t ) R − 1 P = AP + PA T + Q − PC T R − 1 CP, ˙ P ( t 0 ) = P 0 P 0 , Q and R are symmetric, positive definite matrices Assumption 11.1: P ( t ) exists for all t ≥ t 0 and satisfies α 1 I ≤ P ( t ) ≤ α 2 I, α i > 0 Remarks: Assumption 11.1 cannot be checked offline The observer and Riccati equations are solved simultaneously in real time Nonlinear Control Lecture # 31 Nonlinear Observers

Lemma 11.2 There exist positive constants c , k , and λ such that x ( t ) � ≤ ke − λ ( t − t 0 ) , � ˜ x (0) � ≤ c ⇒ � ˜ ∀ t ≥ t 0 ≥ 0 Proof � f ( x, u ) − f (ˆ x, u ) − A ( t )˜ x � = � 1 � � � ∂f x, u ) − ∂f � x � 2 � � 1 ∂x ( σ ˜ x + ˆ ∂x (ˆ x, u ) dσ ˜ x � ≤ 2 L 1 � ˜ � � � 0 � h ( x ) − h (ˆ x ) − C ( t )˜ x � = � 1 � � ∂h � � x ) − ∂h � � 1 x � 2 ∂x ( σ ˜ x + ˆ ∂x (ˆ x ) dσ ˜ x � ≤ 2 L 2 � ˜ � � � 0 Nonlinear Control Lecture # 31 Nonlinear Observers

� � � � ∂h ∂h � � � � � C ( t ) � = ∂x ( x − ˜ x ) � ≤ ∂x (0) � + L 2 ( � x � + � ˜ x � ) � � � � � � x � 2 + k 3 � ˜ x � 3 � ∆(˜ x, x, u ) � ≤ k 1 � ˜ α 1 I ≤ P ( t ) ≤ α 2 I ⇔ α 3 I ≤ P − 1 ( t ) ≤ α 4 I, α i > 0 x T P − 1 ˜ V = ˜ x d dtP − 1 = − P − 1 ˙ PP − 1 Nonlinear Control Lecture # 31 Nonlinear Observers

x T d x T P − 1 ˙ x T P − 1 ˜ ˙ x + ˙ dtP − 1 ˜ V = ˜ ˜ ˜ x + ˜ x x T P − 1 ( A − PC T R − 1 C )˜ = ˜ x x T ( A T − C T R − 1 CP ) P − 1 ˜ + ˜ x x T P − 1 ˙ x T P − 1 ∆ PP − 1 ˜ − ˜ x + 2˜ x T P − 1 ( AP + PA T − PC T R − 1 CP − ˙ P ) P − 1 ˜ = ˜ x x T C T R − 1 C ˜ x T P − 1 ∆ − ˜ x + 2˜ x T ( P − 1 QP − 1 + C T R − 1 C )˜ x T P − 1 ∆ = − ˜ x + 2˜ x � 2 + c 2 k 1 � ˜ x || 3 + c 2 k 2 � ˜ ˙ x || 4 V ≤ − c 1 � ˜ ˙ V ≤ − 1 x � 2 , 2 c 1 � ˜ for � ˜ x � ≤ r Nonlinear Control Lecture # 31 Nonlinear Observers

Example 11.1 x = A 1 x + B 1 [0 . 25 x 2 ˙ 1 x 2 + 0 . 2 sin 2 t ] , y = C 1 x � 0 � � 0 � 1 � � A 1 = , B 1 = , C 1 = 1 0 − 1 − 2 1 Investigate boundedness of x ( t ) P 1 = 1 � 3 � 1 P 1 A 1 + A T 1 P 1 = − I ⇒ 1 1 2 V ( x ) = x T P 1 x Nonlinear Control Lecture # 31 Nonlinear Observers

− x T x + 2 x T P 1 B 1 [0 . 25 x 2 ˙ = 1 x 2 + 0 . 2 sin 2 t ] V −� x � 2 + 0 . 5 � P 1 B 1 � x 2 1 � x � 2 + 0 . 4 � P 1 B 1 �� x � ≤ −� x � 2 + x 2 � x � 2 + 0 . 4 1 = √ √ � x � 2 2 2 √ − 0 . 5 � x � 2 + 0 . 4 for x 2 ≤ √ 2 � x � , 1 ≤ 2 √ √ 2 � T = 2 P − 1 � � � 1 0 1 0 1 √ √ 2 } ⊂ { x 2 Ω = { V ( x ) ≤ 1 ≤ 2 } Nonlinear Control Lecture # 31 Nonlinear Observers

Inside Ω − 0 . 5 � x � 2 + 0 . 4 ˙ √ V ≤ 2 � x � 0 . 4 − 0 . 15 � x � 2 , ≤ ∀ � x � ≥ √ 2 = 0 . 8081 0 . 35 √ (0 . 8081) 2 λ max ( P 1 ) < λ max ( P 1 ) = 1 . 7071 ⇒ 2 ⇒ {� x � ≤ 0 . 8081 } ⊂ Ω ⇒ Ω is positively invariant Design EKF to estimate x ( t ) for x (0) ∈ Ω Nonlinear Control Lecture # 31 Nonlinear Observers

� � 0 1 A ( t ) = x 2 − 1 + 0 . 5ˆ x 1 ( t )ˆ x 2 ( t ) − 2 + 0 . 25ˆ 1 ( t ) � 1 0 � C = Q = R = P (0) = I P = AP + PA T + I − PC T CP, ˙ P (0) = I � p 11 � p 12 P = p 12 p 22 Nonlinear Control Lecture # 31 Nonlinear Observers

˙ ˆ = x 2 + p 11 ( y − ˆ ˆ x 1 ) x 1 ˙ x 2 x 2 ˆ = − ˆ x 1 − 2ˆ x 2 + 0 . 25ˆ 1 ˆ x 2 + 0 . 2 sin 2 t + p 12 ( y − ˆ x 1 ) 2 p 12 + 1 − p 2 ˙ = p 11 11 x 2 p 12 ˙ = p 11 ( − 1 + 0 . 5ˆ x 1 ˆ x 2 ) + p 12 ( − 2 + 0 . 25ˆ 1 ) + p 22 − p 11 p 12 x 2 ˙ = 2 p 12 ( − 1 + 0 . 5ˆ x 1 ˆ x 2 ) + 2 p 22 ( − 2 + 0 . 25ˆ 1 ) p 22 + 1 − p 2 12 Nonlinear Control Lecture # 31 Nonlinear Observers

(a) (b) 1 x 1 1 p 11 x 2 Components of P(t) 0.5 0.8 Estimation Error 0.6 p 22 0 0.4 0.2 −0.5 0 p 12 −0.2 −1 −0.4 0 1 2 3 4 0 1 2 3 4 Time Time Nonlinear Control Lecture # 31 Nonlinear Observers